If $X$ is a compact metric space, $A: X\to X$, is it true that if $a = \inf d(x,Ax),\space x \in X$, then there exists $y \in X$ such that $d(y,Ay) = \inf d(x,Ax)$? If so, why?
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No, it's not true. Let $X = [0,1]$. Define $A$ by $Ax = 1$ for $x\in [0,1)$, and $A1 = 0$. Then $a=0$ but always $d(x,Ax) > 0$. (Obviously, the statement is true if $A$ is continuous.)